How many electrical appliances of the same power (1000 W) can be connected to a 220 V

How many electrical appliances of the same power (1000 W) can be connected to a 220 V network so that the fuse does not blow (the rated current of the fuse is 15A).

Let a certain number of electrical appliances N of the same power P be included in an electrical network with a voltage of U = 220 V. The current strength in each of them will be I₁ = I ₂ = I ₃ = … = P / U, since they are connected in parallel and U = U₁ = U₂ = U₃ =…. The current strength in the unbranched part of the circuit is I = I₁ + I ₂ + I ₃ +… = N ∙ I₁ or I = N ∙ P / U. Then:

N = I ∙ U / P.

From the condition of the problem it is known that the switched on electrical appliances have a power of P = 1000 W, the rated current of the fuse I = 15A (the current at which the fuse does not blow out). We get:

N = (15A ∙ 220 V) / 1000 W;

N = 3.

Answer: so that the fuse does not blow out, 3 electrical appliances of the same power (1000 W each) can be connected to the network.